Chapter 18 - Heat Engines and Refrigerators - Exercises and Problems - Page 549: 13

See the detailed answer below.

a) We know that the work done by the gas per cycle is the enclosed area under the $P−V$ graph. Hence, $$W_{out}={\rm Area}$$ And we can see that the enclosed area under the $PV$ curve is rectangular. Thus, $$W_{out}= (200-100)\times 10^{-6}(400-100)\times 10^3$$ $$W_{out}= \color{red}{\bf 30}\;\rm J$$ We know that $$Q_H=Q_C+W_{out}$$ and we are given $Q_{C}$ in the given graph. Thus, $$Q_H=(90+25)+(30)$$ $$Q_H= \color{red}{\bf 145}\;\rm J$$ --- b) We know that the thermal efficiency is given by $$\eta=\dfrac{W_{out}}{Q_H}=\dfrac{30}{145}$$ $$\eta=\color{red}{\bf 0.21}$$